Is Work Done When Cycling Up a Hill with Constant Velocity?

[Joza]

Work questions!

If you cycle to the top of a hill, without changing your velocity from going from the bottom to the top, is zero work done on you? Is this because with a constant velocity, there is no net force, and work is force times displacement?
But here's my main question. Same scenario as above. but your velocity is slower at the top of the hill. So something has done work to slow your velocity and move you thru a displacement. What has done the work? You? The road?

I did a question on this. It turns out that the work I do by my force on the pedals is the force I apply (my weight) times the vertical distance. Why does y displacement only matter here?
And how come the work that I do is my force (ie weight) times displacement, minus the work done ON ME?

I only recently encountered work, and this is confusing me.

Cheers! ps. I haven't done mgh yet, only kinetic energy

 

[dst]

Work done = force * distance. If you go up a hill at constant v, then you must lose KE which is converted into PE, and the loss in KE = the gain in G.P.E. and gravity is doing the work here (reducing your KE). The original work comes from the force you put up when actually accelerating to that velocity before the hill.

Since gravity only acts in the Y direction and on this hill, is the only force acting to reduce your KE, only Y displacement matters.

 

[CompuChip]

Why does y displacement only matter here?
And how come the work that I do is my force (ie weight) times displacement

That's because the gravitational force (which you must act against to reach the top of the hill) is a conservative force. Perhaps this article will explain a little more, with some examples.

 

[BriK]

Hrm...trick question?

It asks "is zero work done on you"...

If I were to carry a box from point A to point B at a constant velocity, is any work done on the box?

No, because I am applying a vertical force on the box to equal the effects of gravity while I walk horizontally between points...

But, in this question you are changing hieghts on a hill...so it seems work will be done on you because of changes in hieght...

 

[russ_watters]

If you cycle to the top of a hill, without changing your velocity from going from the bottom to the top, is zero work done on you? Is this because with a constant velocity, there is no net force, and work is force times displacement?

Work doesn't say anything about net force being nonzero. In fact, the only time you have a nonzero net force is when you are accelerating. For early study of physics, acceleration makes calculating work/energy more complicated, so the first scenarios you learn to decipher are ones where there is no acceleration - such as this one.

So, in your example, the force is the force applied against gravity and the distance is the distance traveled against gravity.

 

[russ_watters]

Work done = force * distance. If you go up a hill at constant v, then you must lose KE which is converted into PE, and the loss in KE = the gain in G.P.E. and gravity is doing the work here (reducing your KE). The original work comes from the force you put up when actually accelerating to that velocity before the hill.

Since gravity only acts in the Y direction and on this hill, is the only force acting to reduce your KE, only Y displacement matters.

I think you misunderstood. What you are talking about is what happens when you coast up a hill and if you coast up a hill, your v will decrease as you trade KE for PE. The OP described a situation of constant KE, but changing PE.

 

[Shooting Star]

Cheers! ps. I haven't done mgh yet, only kinetic energy

Post this again after you've done mgh; maybe you won't have to post it at all then.